Calculus I · Limits and Continuity · lesson

Choosing Parameters for Piecewise Continuity

Concept

Learning objectives

Set the left-hand expression, right-hand expression, and function value equal at a joining point; solve for unknown parameters.

Choosing Parameters in Piecewise Functions

At a piecewise join $x=a$ , continuity requires

\lim_{x\to a^-}f(x) =f(a) =\lim_{x\to a^+}f(x).

For ordinary polynomial pieces, substitute $x=a$ into both formulas and set the results equal.

Guided walkthrough

One parameter

Find $k$ so that

f(x)= \begin{cases} kx+1,&x<2,\\ x+3,&x\ge2 \end{cases}

is continuous at $2$ .

Show worked solution

Left side at the join:

2k+1.

Right side and function value:

2+3=5.

Set them equal:

2k+1=5.

Subtract $1$ :

2k=4.

Divide by $2$ :

\boxed{k=2}.

Worked example

The parameter appears in both pieces

Find $a$ so that

g(x)= \begin{cases} ax+1,&x<2,\\ x^2-a,&x\ge2 \end{cases}

is continuous at $2$ .

Show worked solution

The left-hand limit is

2a+1.

The right-hand limit and function value are

2^2-a=4-a.

Set them equal:

2a+1=4-a.

Add $a$ to both sides:

3a+1=4.

Subtract $1$ :

3a=3.

Therefore,

\boxed{a=1}.

Worked example

Exam-level: no parameter works

Find $k$ so that

h(x)= \begin{cases} kx+2,&x<1,\\ x^2+k,&x\ge1 \end{cases}

is continuous at $1$ .

Show worked solution

The left-hand expression approaches

k+2.

The right-hand expression and function value equal

1+k.

Continuity would require

k+2=k+1.

Subtracting $k$ gives

2=1,

which is impossible. Therefore,

\boxed{\text{no value of }k\text{ works}}.

A parameter problem is not guaranteed to have a parameter solution. The algebra is allowed to reject the premise, rude though that may seem to a worksheet.

After the explanation

Use the section idea

Reading lens

Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?

Mental model

Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.

Decision

At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.

Common trap

A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.

Check yourself

You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.

Source & rights

Original instruction with traceable references.

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